Answer:
Saddle point: 
Local minimum: 
Local maxima:
, 
Step-by-step explanation:
The function is:

The partial derivatives of the function are included below:




Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.


It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

The solution of the system is (3/8, -3/8).
Let assume that y = 0, the nonlinear system is reduced to a sole expression:

Another solution is (9/8,0).
Now, let consider that x = 0, the nonlinear system is now reduced to this:

Another solution is (0, -9/8).
The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

The second derivatives of the function are:



Then, the expression is simplified to this and each point is tested:

S1: (0,0)
(Saddle Point)
S2: (3/8,-3/8)
(Local maximum or minimum)
S3: (9/8, 0)
(Local maximum or minimum)
S4: (0, - 9/8)
(Local maximum or minimum)
Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:
S2: (3/8,-3/8)
(Local minimum)
S3: (9/8, 0)
(Local maximum)
S4: (0, - 9/8)
(Local maximum)
Saddle point: 
Local minimum: 
Local maxima:
, 